Maths Methods 3/4 Help Thread 2011

[nacho]

when taking the inverse of f(x) = (x – 3)^2/3
and we get to
(y - 3)^2/3 = x-1
why do we take the negative square root next so it is
(y-3) = -(x-1)^3/2
??

[paulsterio]

Was there a domain restriction to begin with?

[paulsterio]

they should, you haven't answered the question properly if you leave +c in

you technically have because "c" can be any number
whether you put +0, +pi, +e, +1000000, +iansdoiwanoinsad or +c, it doesn't matter because it's a constant

[nacho]

Was there a domain restriction to begin with?

oh, yes (negative infinity , 3]
i see now

by the way,
what trial exam company would be most closest to the difficult of vcaa exams?
2010 vcaa, is that as hard as one of the hard exam companies?

[xZero]

you technically have because "c" can be any number
whether you put +0, +pi, +e, +1000000, +iansdoiwanoinsad or +c, it doesn't matter because it's a constant

and you have to state a specific value for c so unless if you write c=... it's not correct

[luffy]

you technically have because "c" can be any number
whether you put +0, +pi, +e, +1000000, +iansdoiwanoinsad or +c, it doesn't matter because it's a constant

and you have to state a specific value for c so unless if you write c=... it's not correct

I don't think VCAA takes marks off for leaving the +c. Mathematically speaking, the +c should always be in there as in real life, finding any ordinary anti-derivative won't be applicable. Last year, we were always taught to include the +c in all cases. Having said that, in the exam, just to be completely safe, I set c to equal 0.

EDIT: If you read the assessment report for Exam 1 2010 (question 2), they even say setting c to a real constant is not necessary.

There you go

[b^3]

you technically have because "c" can be any number
whether you put +0, +pi, +e, +1000000, +iansdoiwanoinsad or +c, it doesn't matter because it's a constant

and you have to state a specific value for c so unless if you write c=... it's not correct

I don't think VCAA takes marks off for leaving the +c. Mathematically speaking, the +c should always be in there as in real life, finding any ordinary anti-derivative won't be applicable. Last year, we were always taught to include the +c in all cases. Having said that, in the exam, just to be completely safe, I set c to equal 0.

Either is fine by VCAA, they won't take marks off.

[nacho]

Two dice are thrown five times. Find the probability that a difference of one occurs at least twice.

Also, a stupid question, but one that gets me all the time;
what is a picture card?
JACK, queen, king AND ace? or no ace?

[pi]

Two dice are thrown five times. Find the probability that a difference of one occurs at least twice.

Also, a stupid question, but one that gets me all the time;
what is a picture card?
JACK, queen, king AND ace? or no ace?

No ace

[b^3]

Two dice are thrown five times. Find the probability that a difference of one occurs at least twice.

Also, a stupid question, but one that gets me all the time;
what is a picture card?
JACK, queen, king AND ace? or no ace?

So a difference of one corresponds to (1,2), (2,1), (2,3), (3,2), (3,4), (4,3), (4,5), (5,4), (5,6), (6,5)
Each of those outcome has a probability of 1/36.
There is 10 outcomes that are success to us, so Pr(difference is 1)=10/36=5/18
The next part is a binomial distribution with p=5/18 and n=5 i.e. X~Bi(5,5/18)
Let X= the number of times our success occurs
Pr(x>=2)=1-Pr(X=0)-Pr(X=1)
=1-⁵C₀(5/18)⁰(1-5/18)⁵-⁵C₁(5/18)¹(1-5/18)⁴ (or you could just do it by calculator, Menu, 5, 5, E)
=0.4256

[nacho]

Two dice are thrown five times. Find the probability that a difference of one occurs at least twice.

Also, a stupid question, but one that gets me all the time;
what is a picture card?
JACK, queen, king AND ace? or no ace?

So a difference of one corresponds to (1,2), (2,1), (2,3), (3,2), (3,4), (4,3), (4,5), (5,4), (5,6), (6,5)
Each of those outcome has a probability of 1/36.
There is 10 outcomes that are success to us, so Pr(difference is 1)=10/36=5/18
The next part is a binomial distribution with p=5/18 and n=5 i.e. X~Bi(5,5/18)
Let X= the number of times our success occurs
Pr(x>=2)=1-Pr(X=0)-Pr(X=1)
=1-⁵C₀(5/18)⁰(1-5/18)⁵-⁵C₁(5/18)¹(1-5/18)⁴ (or you could just do it by calculator, Menu, 5, 5, E)
=0.4256

thanks, i just realised i substracted Pr(X=2) as  well

[nacho]

 for continuous distributions, im not sure if i understand what the x and y axis represent?
before i thought x axis were the values that could be taken (eg weight = 43kg) would be (43, y) on the x-axis
but then somtimes there are negative values?
and what about y-axis?
thanks

[duquesne9995]

for continuous distributions, im not sure if i understand what the x and y axis represent?
before i thought x axis were the values that could be taken (eg weight = 43kg) would be (43, y) on the x-axis
but then somtimes there are negative values?
and what about y-axis?
thanks

the x-axis can take negative values too, depending on the context of the question, obviously it can't have negative values of weight but for another problem, it can be negative

the y values are not as relevant to the probability, because the probability is the area under the curve
they are only important when looking at the mode of the distribution, where the mode is the x-value which gives the greatest y-value, but you wouldn't need to find the y-value

my question:
in Exam 1, when differentiating, are we allowed to leave answers in sec^2(x) or do we have to write 1/cos^2(x)
Thanks =]

[duquesne9995]

e^2x +be^x +1 = 0 where b is a real constant
for what values of b will x have no real solutions?
thx in advance

[thushan]

sec^2 (x) is fine.